Let the principal be \(x\) rupees. In 4 years, the amount becomes \(2x\) under compound interest. \[ \therefore 2x = x(1 + \frac{r}{100})^4 \Rightarrow (1 + \frac{r}{100})^4 = 2 \tag{i} \] Now, suppose the same principal \(x\) becomes \(4x\) in \(n\) years. \[ \therefore 4x = x(1 + \frac{r}{100})^n \Rightarrow (1 + \frac{r}{100})^n = 4 \Rightarrow (1 + \frac{r}{100})^n = 2^2 \Rightarrow (1 + \frac{r}{100})^n = [(1 + \frac{r}{100})^4]^2 \Rightarrow (1 + \frac{r}{100})^n = (1 + \frac{r}{100})^8 \Rightarrow n = 8 \] ∴ At a fixed compound interest rate, if a sum of money doubles in 4 years, it will become four times in 8 years.